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Mean and Variance of Continuous Random Variable


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$begingroup$


I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac{1}{(1+e^{-y})}$.



Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.



Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?










share|cite|improve this question









New contributor




EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



















    2












    $begingroup$


    I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac{1}{(1+e^{-y})}$.



    Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.



    Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?










    share|cite|improve this question









    New contributor




    EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2





      $begingroup$


      I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac{1}{(1+e^{-y})}$.



      Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.



      Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?










      share|cite|improve this question









      New contributor




      EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac{1}{(1+e^{-y})}$.



      Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.



      Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?







      variance mean






      share|cite|improve this question









      New contributor




      EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Noah

      3,6011417




      3,6011417






      New contributor




      EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 5 hours ago









      EBuschEBusch

      112




      112




      New contributor




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      New contributor





      EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          What makes you think you did something wrong?



          begin{align}
          & Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
          text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
          = {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
          end{align}

          and therefore
          $$
          Pr(Yle y) = Pr(Y ge -y).
          $$

          So this distribution is symmetric about $0.$



          Therefore, if the expected value exists, it is $0.$



          You can also show that the density function is an even function:
          begin{align}
          f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
          f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
          end{align}

          Since the density is an even function, the expected value must be $0$ if it exists.



          The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Comment:



            Setting what I take to be your CDF equal to $U sim mathsf{Unif}(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
            observations as shown below.



            Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
            seems to fit pretty well.



            set.seed(1019)  # for reproducibility
            u = runif(10^6); x = -log(1/u - 1)

            hist(x, prob=T, br=100, col="skyblue2")
            curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")


            enter image description here



            I don't pretend that this is a 'worked answer' to your problem, but
            I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.






            share|cite|improve this answer











            $endgroup$














              Your Answer





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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              What makes you think you did something wrong?



              begin{align}
              & Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
              text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
              = {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
              end{align}

              and therefore
              $$
              Pr(Yle y) = Pr(Y ge -y).
              $$

              So this distribution is symmetric about $0.$



              Therefore, if the expected value exists, it is $0.$



              You can also show that the density function is an even function:
              begin{align}
              f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
              f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
              end{align}

              Since the density is an even function, the expected value must be $0$ if it exists.



              The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                What makes you think you did something wrong?



                begin{align}
                & Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
                text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
                = {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
                end{align}

                and therefore
                $$
                Pr(Yle y) = Pr(Y ge -y).
                $$

                So this distribution is symmetric about $0.$



                Therefore, if the expected value exists, it is $0.$



                You can also show that the density function is an even function:
                begin{align}
                f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
                f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
                end{align}

                Since the density is an even function, the expected value must be $0$ if it exists.



                The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  What makes you think you did something wrong?



                  begin{align}
                  & Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
                  text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
                  = {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
                  end{align}

                  and therefore
                  $$
                  Pr(Yle y) = Pr(Y ge -y).
                  $$

                  So this distribution is symmetric about $0.$



                  Therefore, if the expected value exists, it is $0.$



                  You can also show that the density function is an even function:
                  begin{align}
                  f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
                  f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
                  end{align}

                  Since the density is an even function, the expected value must be $0$ if it exists.



                  The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$






                  share|cite|improve this answer









                  $endgroup$



                  What makes you think you did something wrong?



                  begin{align}
                  & Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
                  text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
                  = {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
                  end{align}

                  and therefore
                  $$
                  Pr(Yle y) = Pr(Y ge -y).
                  $$

                  So this distribution is symmetric about $0.$



                  Therefore, if the expected value exists, it is $0.$



                  You can also show that the density function is an even function:
                  begin{align}
                  f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
                  f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
                  end{align}

                  Since the density is an even function, the expected value must be $0$ if it exists.



                  The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Michael HardyMichael Hardy

                  3,9951430




                  3,9951430

























                      0












                      $begingroup$

                      Comment:



                      Setting what I take to be your CDF equal to $U sim mathsf{Unif}(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
                      observations as shown below.



                      Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
                      seems to fit pretty well.



                      set.seed(1019)  # for reproducibility
                      u = runif(10^6); x = -log(1/u - 1)

                      hist(x, prob=T, br=100, col="skyblue2")
                      curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")


                      enter image description here



                      I don't pretend that this is a 'worked answer' to your problem, but
                      I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Comment:



                        Setting what I take to be your CDF equal to $U sim mathsf{Unif}(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
                        observations as shown below.



                        Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
                        seems to fit pretty well.



                        set.seed(1019)  # for reproducibility
                        u = runif(10^6); x = -log(1/u - 1)

                        hist(x, prob=T, br=100, col="skyblue2")
                        curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")


                        enter image description here



                        I don't pretend that this is a 'worked answer' to your problem, but
                        I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Comment:



                          Setting what I take to be your CDF equal to $U sim mathsf{Unif}(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
                          observations as shown below.



                          Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
                          seems to fit pretty well.



                          set.seed(1019)  # for reproducibility
                          u = runif(10^6); x = -log(1/u - 1)

                          hist(x, prob=T, br=100, col="skyblue2")
                          curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")


                          enter image description here



                          I don't pretend that this is a 'worked answer' to your problem, but
                          I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.






                          share|cite|improve this answer











                          $endgroup$



                          Comment:



                          Setting what I take to be your CDF equal to $U sim mathsf{Unif}(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
                          observations as shown below.



                          Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
                          seems to fit pretty well.



                          set.seed(1019)  # for reproducibility
                          u = runif(10^6); x = -log(1/u - 1)

                          hist(x, prob=T, br=100, col="skyblue2")
                          curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")


                          enter image description here



                          I don't pretend that this is a 'worked answer' to your problem, but
                          I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 19 mins ago

























                          answered 4 hours ago









                          BruceETBruceET

                          6,3531721




                          6,3531721






















                              EBusch is a new contributor. Be nice, and check out our Code of Conduct.










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